Finite Element Analysis - Thermomechanics of Solids Part 14 pps

Torsion and Buckling 14.1 TORSION OF PRISMATIC BARS Figure 14.1 illustrates a member experiencing torsion.. It follows that 14.2 BUCKLING OF BEAMS AND PLATES 14.2.1 E ULER B UCKLING OF B

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Torsion and Buckling

14.1 TORSION OF PRISMATIC BARS

Figure 14.1 illustrates a member experiencing torsion The member in this case is cylindrical with length L and radius r0 The base is fixed, and a torque is applied at the top surface, which causes the member to twist The twist at height z is θ(z), and

at height L, it is θ0 Ordinarily, in the finite-element problems so far considered, the displacement is the basic unknown It is approximated by an interpolation model, from which an approximation for the strain tensor is obtained Then, an approximation for the stress tensor is obtained using the stress-strain relations The nodal displacements are solved by an equilibrium principle, in the form of the Principle of Virtual Work In the current problem, an alternative path is followed in which stresses or, more precisely, a stress potential, is the unknown The strains are determined from the stresses However, for arbitrary stresses satisfying equilibrium, the strain field may not be compatible The compatibility condition (see Chapter 4) is enforced, furnishing

FIGURE 14.1 Twist of a prismatic rod.

14

section before twist

section after twist

z T

θ 0

r0

θ

φ x

z

y

L x

0749_Frame_C14 Page 181 Wednesday, February 19, 2003 5:41 PM

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182 Finite Element Analysis: Thermomechanics of Solids

a partial differential equation known as the Poisson Equation A variational argument

is applied to furnish a finite-element expression for the torsional constant of the section

For the member before twist, consider points X and Y at angle φ and at radial position r Clearly, X=r cosφ and Y=r sinφ Twist induces a rotation through angle

θ(z), but it does not affect the radial position Now, x=r cos(φ+θ), y=r sin(φ+θ) Use of double-angle formulae furnishes the displacements, and restriction to small angles θ furnishes, to first order,

(14.1)

It is also assumed that torsion does not increase the length of the member, which

is attained by requiring that axial displacement w only depends on X and Y The quantity w(X, Y) is called the warping function

It is readily verified that all strains vanish except E xz and E yz, for which

(14.2)

Equilibrium requires that

(14.3)

The equilibrium relation can be identically satisfied by a potential function y

for which

(14.4)

We must satisfy the compatibility condition to ensure that the strain field arises from a displacement field that is unique to within a rigid-body translation and rotation (Compatibility is automatically satisfied if the displacements are considered the unknowns and are approximated by a continuous interpolation model Here, the stresses are the unknowns.) From the stress-strain relation,

(14.5)

Compatibility (integrability) now requires that , furnishing

(14.6)

u= −Yθ, v=Xθ

w

∂ − ∂∂







∂

∂ + ∂∂





1 2

∂

S x

S y

xz yz

0

S

xz= ∂ yz

∂

1 2

∂

∂ ∂ 2 w = ∂ ∂∂2

x y

w

y x

− ∂

∂

∂ +







 +∂∂ − ∂∂ −





 =

d

d dz

1 2

1

µ

,

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Torsion and Buckling 183

which in turn furnishes Poisson’s Equation for the potential function y:

(14.7)

For boundary conditions, assume that the lateral boundaries of the member are unloaded The stress-traction relation already implies that τx= 0 and τy= 0 on the lateral boundary S For traction τz to vanish, we require that

(14.8)

Upon examining Figure 14.2, it can be seen that n x=dy/ds and n y=−dx/ds, in which s is the arc length along the boundary at z Consequently,

(14.9) Now, on and therefore ψ is a constant, which can, in general, be taken

as zero

We next consider the total torque on the member Figure 14.3 depicts the cross section at z The torque on the element at x and y is given by

(14.10)

FIGURE 14.2 Illustration of geometric relation.

ψ

n

nx = cos χ = dy/ds

ny = sin χ = dx/ds

∂

∂ + ∂∂ = −

2 2 2

d

dz .

τz =n S x xz+n S y yz = 0 on S

τ

ψ

dy

ds S

dx

ds S dy

ds y

dx

ds x d

ds

∂ +

∂

=

S,d dsψ = 0,

dT xS dxdy yS dxdy

x d

dx y

d

dy dxdy







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Integration furnishes

(14.11)

Application of the divergence theorem to the first term leads to ∫ψ[xn x+yn y]ds, which vanishes since y vanishes on S Finally,

(14.12)

We apply variational methods to the Poisson Equation, considering the stress-potential function y to be the unknown Now,

(14.13)

FIGURE 14.3 Evaluation of twisting moment.

z

x

y

yz

sxz dy

dx y

d

dy dxdy

d x dx

d y dy

dx dx

dy

dy dxdy x



  −  + 









= − ∇⋅





∫

ψ

2

T=2∫ψdxdy

δψ[∇ ⋅ ∇ +ψ µθ′] =

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Integration by parts, use of the divergence theorem, and imposition of the

“constraint” y= 0 on S furnishes

(14.14)

The integrals are evaluated over a set of small elements In the e th element,

approximate ψ as in which νT is a vector with dimension (number of

rows) equal to the number of nodal values of ψ The gradient ∇ψ has a corresponding

interpolation model in which ββββT is a matrix The finite-element

counterpart of the Poisson Equation at the element level is written as

(14.15)

and the stiffness matrix should be nonsingular, since the constraint y= 0 on S has

already been used It follows that, globally, The torque satisfies

(14.16)

In the theory of torsion, it is common to introduce the torsional constant J, for

which T= 2µJθ′ It follows that

14.2 BUCKLING OF BEAMS AND PLATES

14.2.1 E ULER B UCKLING OF B EAM C OLUMNS

14.2.1.1 Static Buckling

Under in-plane compressive loads, the resistance of a thin member (beam or plate)

can be reduced progressively, culminating in buckling There are two equilibrium

states that the member potentially can sustain: compression only, or compression

with bending The member will “snap” to the second state if it involves less “potential

energy” than the first state The notions explaining buckling are addressed in detail in

subsequent chapters For now, we will focus on beams and plates, using classical

equations in which, by retaining lowest-order corrections for geometric nonlinearity,

in-plane compressive forces appear

∫ δψ ψdxdy ∫δψ µθ2 dxdy

νTT( , )x yψψ ηηe e,

∇ =ψ ββTT( , )x yψψ ηηe e,

K

f

T

T e

e

T e

e T

dxdy

x y dxdy

( )

( , )

ηη

ψψ ββ ββ ψψ ψψ

=

∫

2µθ

ν

ηηg

=2µθ ′ T − T

( ) ( )

T

=

∫

( ) ( ) ( ) − ( )

2

4

ψ

µθ

ηηg

f

J=2fT( )g TKT( )g−1fT( )g

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For the beam shown in Figure 14.4, the classical Euler buckling equation is

(14.17)

and P is the axial compressive force The interpolation model for w(x) is recalled

as w(x) = ϕϕϕϕT

(x)ΦΦγγγγ Following the usual variational procedures (integration by parts) furnishes

(14.18)

At x = 0, both δ w and −δw′ vanish, while the shear force V and the bending

moment M are identified as V = −EIw′′′ and M = −EIw′′ The “effective shear force”

Q is defined as Q = −Pw′ − EIw′′′.

For the specific case illustrated in Figure 14.3, for a one-element model, we can

use the interpolation formula

(14.19)

The mass matrix is shown, after some algebra, to be

(14.20)

FIGURE 14.4 Euler buckling of a beam column.

z

E,I,A,L, ρ y

x

Q0

M0P

EIw iv+Pw′′ +ρ ˙˙Aw=0,

δ ρw Aw dx˙˙ δ ˙˙, ( )xρA ( )x dV

∫ → γγT γγ =ΦΦ ϕϕT∫ ϕϕT Φ

iv

w L

( ) ( )

=





















−

2 1











13

35 21011 11

210 1051

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Similarly,

(14.21)

The governing equation is written in finite-element form as

(14.22)

In a static problem, the solution has the form

(14.23)

in which cof denotes the cofactor, and γγγγ → ∞ for values of which render

det(K2

14.2.1.2 Dynamic Buckling

In a dynamic problem, it may be of interest to determine the effect of P on the

resonance frequency Suppose that f(t) = f0 exp(i ωt), in which f0 is a known vector The displacement function satisfies γγγγ(t) = γγγγ0 exp(i ωt), in which the amplitude vector

γγγγ0 satisfies

(14.24)

Resonance occurs at a frequency ω0, for which

(14.25)

resonance frequency is reduced by the presence of P and vanishes precisely at the critical value of P.





















∫

L

T

6

5 101 1

10 152

,

E

,

I L

P

Q M

0











˙˙ ,

γγ = 0

γγ =

−









−









cof

det

,

f

PL I PL I

2

E

PL I

2

E

I

2

EI L

P

2



L

P

2



0

1 2

1 2 3

ρALK−/ E K − K K−/

L P L

ω02

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14.2.1.3 Sample Problem: Interpretation of Buckling Modes

Consider static buckling of a clamped-clamped beam, as shown in Figure 14.5

This configuration can be replaced with two beams of length L, for which the right beam experiences shear force V1 and bending moment M1, while the left beam

experiences shear force V0− V1 and bending moment M0− M1 The beam on the right is governed by

(14.26)

The governing equation is written in finite-element form as

(14.27)

Consider the symmetric case in which M0= 0, with the implication that w′(L) = 0.

The equation reduces to

(14.28)

from which we obtain the critical buckling load given by P1= 10EI/L2

FIGURE 14.5 Buckling of a clamped-clamped beam.

M0

V0

P





















∫

L

T

6

5 101 1

10 152

,

EI L

L

P L

L

6

5 101 1

10 152













−

























γγ f=

f=



 



− ′









V M

w L

1

,

( ) ( ) γγ

5

EI L

P

−



  ( )= ,

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Next, consider the antisymmetric case in which V0= 0 and w(L) = 0 The

coun-terpart of Equation 14.28 is now

(14.29)

and P2= 30EI/L2

If neither the constraint of symmetry nor axisymmetry is applicable, there are two critical buckling loads, to be obtained in Exercise 2 as and These values are close enough to the symmetric and antisymmetric cases to suggest an interpretation

of the two buckling loads as corresponding to the two “pure” buckling modes Compare the obtained values with the exact solution, assuming static

condi-tions Consider the symmetric case Let w(x) = w c (x) + w p (x), in which w c (x) is the characteristic solution and w p (x) is the particular solution reflecting the per-turbation From the Euler buckling equation demonstrated in Equation 14.17, w c (x) has a general solution of the form w c (x) = α + βx + γ cosκx + δ sinκx, in which

κ = Now, w = −w′ = 0 at x = 0, −w′(L) = 0, and EIw′′′(L) = V1, expressed

as the conditions

(14.30)

or otherwise stated

(14.31)

For the solution to “blow up,” it is necessary for the matrix B to be singular,

which it is if the corresponding homogeneous problem has a solution Accordingly,

we seek conditions under which there exists a nonvanishing vector z, for which Bz = 0 Direct elimination of α and β furnishes α = −γ and β = −κδ The remaining coefficients must satisfy

(14.32)

EI



 (− ′( ))= ,

27 8 EI L2 8 9 EI L2

PL2/EI

0

w w

p

( ) ( )

−

− ′

















=

−













=







w w

w L

p

( ) ( ) ( ) ( )

,

0 0

0

1

E

α β γ δ



























 



 =



 





γ δ

0 0

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A nonvanishing solution is possible only if the determinant vanishes, which reduces to sinκL = 0 This equation has many solutions for kL, including kL = 0 The lowest nontrivial solution is kL = p, from which P crit= π2

EI/L2= 9.87 EI/L2

Clearly, the symmetric solution in the previous two-element model (P crit = 10 EI/L2

) gives an accurate result

For the antisymmetric case, the corresponding result is that tanκL = κL The lowest meaningful root of this equation is kL = 4.49 (see Brush and Almroth, 1975),

giving P crit = 20.19 EI/L2

Clearly, the axisymmetric part of the two-element model is not as accurate, unlike the symmetric part This issue is addressed further in the subsequent exercises

Up to this point, it has been implicitly assumed that the beam column is initially perfectly straight This assumption can lead to overestimates of the critical buckling

load Consider a known initial distribution w0(x) The governing equation is

(14.33)

or equivalently,

(14.34)

The crookedness is modeled as a perturbation Similarly, if the cross-sectional properties of the beam column exhibit a small amount of variation, for example,

EI(x) = EI0[1 + ϑ sin(πx/L)], the imperfection can also be modeled as a perturbation

14.2.2 E ULER B UCKLING OF P LATES

The governing equation for a plate element subject to in-plane loads is

(14.35)

(see Wang 1953), in which the loads are illustrated as shown in Figure 14.6 The usual

FIGURE 14.6 Plate element with in-plane compressive loads.

d

dx I

d

2 2 2

2

d

dx I

d

dx w

d

dx I

d

dx w

2 2 2 2

2 2

2 2 2

2

Eh

w

x y

2 2

2 2 2

2

∂

∂ ∂ = ν

z

x

Px

Pyx

Py

Pxy h

y

Trang 11

variational methods furnish, with some effort,

(14.36)

in which W = ∇∇T

w (a matrix!) In addition,

(14.37)

For simplicity’s sake, assume that from which we can obtain the form

(14.38)

We also assume that the secondary variables (n ⋅ ∇)∇2

w, (n ⋅ ∇)∇w, and also are prescribed on S.

These conditions serve to obtain

(14.39)

and f reflects the quantities prescribed on S.

As illustrated in Figure 14.7, we now consider a three-dimensional loading space

in which P x , P y , and P xy correspond to the axes, and seek to determine a surface in the space of critical values at which buckling occurs In this space, a straight line

δw∇w dA= δw ⋅∇ ∇w dS− δ∇ ⋅w ⋅∇ ∇w dS+ trδ dA

δ

w

w dA

x

y

∂

2 2 2 2

2







 





∫

p

P

+

























∂

xy y

w

1 2 1 2

1 2 1 2 ,

w x y( , )= ϕϕ Φb2Φ γγb2 b2

T

∇ =



 



w w

x

y

ββ Φ1 2T Φ γγ2 2 ββ Φ2 2T Φ γγ2 2

W

w

x

w y w

x

w y

∂

+













1 2 1

2

[Kb21−Kb22]γγb2=f K

K

E

d

21

2

12 1

=

−

=

∫

h

A

Φ ββ Pββ VΦ

Trang 12

emanating from the origin represents a proportional loading path Let the load

intensity, λ, denote the distance to a given point on this line By analogy with spherical coordinates, there exist two angles, θ and φ, such that

(14.40)

Now,

(14.41)

For each pair (q, f), buckling occurs at a critical load intensity, λ crit(θ, φ), satisfying

(14.42)

A surface of critical load intensities, λcrit(θ, φ), can be drawn in the loading space shown in Figure 14.7 by evaluating λcrit(θ, φ) over all values of (q, f) and discarding values that are negative

FIGURE 14.7 Loading space for plate buckling.

Pxy

Py

Px

λ φ θ

P x =λcos cos ,θ φ P y =λsin cos ,θ φ P xy =λsinφ

P

=

=











∫

ˆ ( , )

cos

det[Kb21−λ θ φcrit( , ) ˆKb22]=0

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Torsion and Buckling 193 14.3 EXERCISES

1 Consider the triangular member shown to be modeled as one finite ele-ment Assume that

Find K T , f T , and the torsional constant T.

2 Find the torsional constant for a unit square cross section using two triangular elements

3 Derive the matrices K0, K1, and K2 in Equations 14.20 and 14.21

4 Compute the two critical values in Equation 14.23

5 Use the four-element model shown in Figure 14.5, and determine how much improvement, if any, occurs in the symmetric and antisymmetric cases

6 Consider a two-element model and a four-element model of the

simple-simple case shown in the following figure Compare P crit in the symmetric and antisymmetric cases with exact values

Triangular shaft cross section.

ψ

ψ ψ ψ

=





























−

1

1 1 1

1 1

2

3

y

x

(2,3)

(3,2)

(1,1)

V 0

M 0

P

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